Question: $f\,^{\prime}(x)=5e^x$ and $f(7)=40+5e^7$. $f(0) = $
Solution: Finding $f(x)$ We have $f'(x)=5e^x$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int (5e^x)\,dx \\\\ & = {5e^x} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(7)=40+5e^7$. Here's what we get when we plug in $7$ : $\begin{aligned}f(7)&={5e^{7}} {+ C} \end{aligned}$ We are given that this must equal $40+5e^7$ : $40+5e^7 = {5e^{7}} {+ C}$ Solving the equation gives us ${C=40}$. Finding $f(0)$ Now, we have that $f(x)={5e^x} {+ 40}$. Let's find $f(0)$ by plugging in $0$ : $\begin{aligned}f(0)&=5e^0 + 40\\\\ &=45 \end{aligned}$ The answer $f(0) = 45$